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MATHS 2 Marks Questions

Mean, Median, Mode of Grouped Data · MP Board (MPBSE) STD 10 (Madhya Pradesh - English Medium). 21 questions.

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21 questions · timed · auto-graded

Question 12 Marks
Following is the distribution of marks of 70 students in a periodical test:
Marks
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Number of students
3
11
28
48
70
Draw a cumulative frequency for the above data.
Answer
Marks
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Number of students
3
11
28
48
70
We plot the points (10,3), (20,11), (30,28), (40,48) and (50,70). to get the cumulative frequency curve as follows:
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Question 22 Marks
A data has 25 observations arranged in a descending order. Which observation represents the median?
Answer
Number of observation $= n = 25 (odd)$
Hence, median = value of $\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}$ observation
= Value of $\Big(\frac{25+1}{2}\Big)^\text{th}$ observation
$=$ Value of $13^{\text {th }}$ observation
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Question 32 Marks
The distributions X and Y with total number of observations 36 and 64, and mean 4 and 3 respectively are combined. What is the mean of resulting distribution X + Y?
Answer
For drsmbunoo X:
Total number of observations = 36
Mean = 4
⇒ Sum ofobservabons = 36 × 4 = 144
For drstnbuneo Y:
Total number of observaoons = 64
Mean =3
⇒ Sum of observations = 36 × 4 = 192
For distribution X + Y:
Total number of observations = 36 + 64 = 100
Sum of observeuoos = 144 + 192 = 336
$\Rightarrow\text{Mean}=\frac{336}{100}=3.36$
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Question 42 Marks
Write the median class of the following distribution:
Class
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
4
4
8
10
12
8
4
Answer
Class interval
Frequency
Cumulative frequency
0-10
4
4
10-20
4
8
20-30
8
16
30-40
10
26
40-50
12
38
50-60
8
46
60-70
4
50
Here, $\text{N}=50\Rightarrow\frac{\text{N}}{2}=25$
The cumulative frequency just greater 111an 25 is 26.
Hence, median dass is 30- 40.
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Question 52 Marks
Find the mean of the following frequency distribution using step-deviation method:
Class
84-90
90-96
96-102
102-108
108-114
114-120
Frequency
15
22
20
18
20
25
Answer
Class interval
Frequency $f_i$
Mid values $x_i$
 $\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{A}}{\text{h}}=\frac{\text{x}_i-99}{6}$
$f_i \times u_i$
84-90
15
87
 -2
-30
90-96
 22
93
 -1
-22
96-102
 20
99 = A
 0
0
102-108
18
105
 1
18
108-114
20
111
 2
40
114-120 25 117 3 75
  $\sum\text{f}_\text{i}=120$     $\sum\text{f}_\text{i}\text{u}_\text{i}=81$
Thus, $\text{A}=99,\ \text{h}=6,\ \sum\text{f}_\text{i}=120$ and $\sum\text{f}_\text{i}\text{u}_\text{i}=81$
Mean $=\text{A}+\Big\{\text{h}\times\frac{\sum\text{f}_\text{i}\text{u}\text{i}}{\sum\text{f}_\text{i}}\Big\}$
$=99+\Big\{6\times\frac{81}{120}\Big\}$
$=99+4.05$
$=103.05$
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Question 62 Marks
The median value for the following frequency distribution is $35$ and the sum of the all frequencies is $170$. Using the formula for median, find the missing frequencies.
Class
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
10
20
?
40
?
25
15
Answer
Class interval
Frequency $f_i$
Cumulative frequency cf
0-10
10
10
10-20
20
30
20-30
$f_i$
$30 + f_i​​​​​​​$​​​​​​​
30-40
40
$70 + f_i​​​​​​​$
40-50
$f_2​​​​​​​$
$70 + f_1 + f_2​​​​​​​$
50-60
25
$95 + f_1 + f_2​​​​​​​$
60-70
15
$110 + f_1 + f_2 = 170$
Here, $\text{N}\Rightarrow\frac{\text{N}}{2}=85$
Median is 35, wgich lies in the class 30-40.
Hencw, median class is 30-40.
$\therefore\text{l}=30,\ \text{h}=10,\ \text{f}=40,\ \text{cf}=$ cf of preceding class $= 30 + f_1​​​​​​​$​​​​​​​
Now, Median $=\text{l}+\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$\therefore35=30+\Big\{10\times\frac{85-30-\text{f}_1}{40}\Big\}$
$\therefore35=30+\Big\{10\times\frac{55-\text{f}_1}{40}\Big\}$
$\therefore5=\frac{55-\text{f}_1}{4}$
$\therefore20=55-\text{f}_1$
$\therefore\text{f}_1=55-20=35$
Now, $100+\text{f}_1+\text{f}_2=170$
$\Rightarrow110+35+\text{f}_2=170$
$\Rightarrow\text{f}_2=170-145=25$
Hence, the missing frequencies are 35 and 25.
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Question 72 Marks
For the following distribution draw a 'less than type' ogive and from the curve find the median.
Marks obtained
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
Less than 80
Less than 90
Less than 100
Number of students
2
7
17
40
60
82
85
90
100
Mean, Median, Mode of grouped Data, Cumulative Frequency Graph and Ogive.
Answer
Marks obtained
Less than 20
 Less than 30 Less than 40 Less than 50
Less than 60
 Less than 70 Less than 80 Less than 90 Less than 100
Number of student
2
 7 17 40 60 82 85 90 100
We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90), (100, 100) to get the cumulative frequency curve as follows:

Here, $\text{N}=100\Rightarrow\frac{\text{N}}{2}=50$
At y = 50, affix A.
Through A, draw a horizontal line is drawn which meets OX at M.
OM = 56
Hence, median = 56
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Question 82 Marks
The maximum bowling speeds (in km/h) of 33 players at a cricket coaching centre are given below:
Speed in km/h
85-100
100-115
115-130
130-145
No. of players
10
4
7
9
Calculate the median bowling speed.
Answer
Class interval
Frequency
Cumulative Frequency
85-100
10
10
100-115
4
14
115-130
7
21
130-145
9
30
Here, $\text{N}=30\Rightarrow\frac{\text{N}}{2}=15$
The cumulative frequency just greater than 15 is 21.
$\therefore\text{l}=115,\ \text{h}=15,\ \text{f}=7,\ \text{cf}=$ cf of preceding class = 14
Now, Median $=\text{l}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$=115+\Big\{15\times\frac{(15-14)}{7}\Big\}$
$=115+\Big\{15\times\frac{1}{7}\Big\}$
$=115+2.1$
$=117.1$
Thus, the median bowling speed is 117.1km/hr
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Question 92 Marks
Find the missing frequencies $f_1$ and $f_2$ in the table in given below, it is being given that the mean of the given frequency distribution is $50.$
Class
0-20
20-40
40-60
60-80
80-100
Total
Frequency
17
$f_1$
32
$f_2​​​​​​​$
19
120
Answer
Class
Frequency
Mid values ($x_i$​​​​​​​)
$(f_i \times x_i)$
0-20
17
10
170
20-40
$f_1$​​​​​​​
30
$30 f_1​​​​​​​$
40-60
32
50
1600
60-80
$52 - f_1​​​​​​​$
70
$3640 - 70 f_1​​​​​​​$
80-100
19
90
1710
Now, Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow50=\frac{3480+\text{f}_1+70\text{f}_2}{120}$
$\Rightarrow6000=3480+30\text{f}_1+70\text{f}_2$
$\Rightarrow30\text{f}_1+70\text{f}_2=2520$
$\Rightarrow3\text{f}_1+7\text{f}_2=252\ ...(\text{i})$
Now, $68+\text{f}_1+\text{f}_2=120$
$\Rightarrow\text{f}_1+\text{f}_2=52$
$\Rightarrow\text{f}_1=52-\text{f}_2$
Substituting in (i), we have
$3(52-\text{f}_2)+7\text{f}_2=252$
$\Rightarrow156-3\text{f}_2+7\text{f}_2=252$
$\Rightarrow4\text{f}_2=96\Rightarrow\text{f}_2=24$
$\Rightarrow\text{f}_2=52-24=28$
Thus, the missing frequencies are $\text{f}_1=28$ and $\text{f}_2=24.$
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Question 102 Marks
In a class test, 50 students obtained marks as follows:
Marks obtained
0-20
20-40
40-60
60-80
80-100
Number of students
4
6
25
10
5
Find the modal class and the median class.
Answer
Oass having maximum frequency is the modal dass.
Here, maximum frequency. 25
Hence, the modal dass rs 40-60
Class interval
Number of students
Cumulative frequency
0-20
4
4
20-40
6
10
40-60
25
35
60-80
10
45
80-100
5
50
Here, $\text{N}=50\Rightarrow\frac{\text{N}}{2}=25$
The cumulative frequency just greater than 25 is 35.
Hence, the median class is 40- 60.
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Question 112 Marks
Find the median of the following frequency distribution:
Marks
0-10
10-20
20-30
30-40
40-50
Number of students
6
16
30
9
4
Answer
Class interval
Frequency
Cumulative frequency
0-10
6
6
10-20
16
22
20-30
30
52
30-40
9
61
40-50
4
65
Here, $\text{N}=65\Rightarrow\frac{\text{N}}{2}=32.5$
The cumulative frequency just greater than 32.5 is 52.
Hence, median class is 20-30.
$\therefore\text{l}=20,\ \text{h}=10,\ \text{f}=30,\ \text{cf}=$ cf of preceding class = 14
Now, Median $=\text{l}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$=20+\Big\{10\times\frac{10.5}{30}\Big\}$
$=20+3.5$
$=23.5$
thus, the median of the data is 23.5
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Question 122 Marks
The monthly pocket money of 50 students of a class are given in the following distribution:
Monthly pocket money (in Rs.)
0-50
50-100
100-150
150-200
200-250
250-300
Number of students
2
7
8
30
12
1
Find the modal class and also give class mark of the modal class.
Answer
Clss having maximum frequency is the modal class.
Here, maximum frequency = 30
Hence, the modal cass is 150-200
$\therefore$ Class-mark of the modal cass $=\frac{150+200}{2}=\frac{350}{2}=175$
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Question 132 Marks
Find the class marks of classes 10-25 and 35-55.
Answer
Class- mark of the 10-25 $=\frac{10+25}{2}=\frac{35}{2}=17.5$
Class-mark of class 35-55 $\frac{35+55}{2}=\frac{90}{2}=45$
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Question 142 Marks
For a certain distribution, mode and median were found to be 1000 and 1250 respectively. Find mean for this distribution using an empirical relation.
Answer
Mode = 1000
Medion = 1250
Empirical Relationship:
Mode = 3Median - 2Mean
$\Rightarrow\text{Mean}=\frac{3\text{Meduian}-\text{Mode}}{2}$
$=\frac{3\times1250-1000}{2}$
$=\frac{3750-1000}{2}$
$=\frac{2750}{2}$
$=1375$
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Question 152 Marks
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?
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Question 162 Marks
Write the lower limit of the modal class of the following frequency distribution?
Age (in years)
0-10
10-20
20-30
30-40
40-50
50-60
Number of patients
16
13
6
11
27
18
Answer
Class having maximum frequency is the modal class.
Here, maximum frequency = 27
Hence, the modal class is 40-50.
Thus, the lower limit of the modal class is 40.
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Question 172 Marks
If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.
Answer
$\text{Mean}=\frac{\text{Sum of given observations}}{\text{Total number of observations}}$$\therefore11=\frac{\text{x}+(\text{x}+2)+(\text{x}+4)+(\text{x}+6)+(\text{x}+8)}{5}$
$\Rightarrow55=5\text{x}+20$
$\Rightarrow5\text{x}=55-20$
$\Rightarrow5\text{x}=35$
$\Rightarrow\text{x}=\frac{35}{5}$
$\Rightarrow\text{x}=7$
Hence, the value of x is 7.
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Question 182 Marks
Find the mean of the following frequency distribution:
Class
1-3
3-5
5-7
7-9
Frequency
9
22
27
18
Answer
We have,
Class interval
Frequency $f_i$
Mid-value $x_i$
$f_i \times x_i$
1-3
9
2
18
3-5
22
4
88
5-7
27
6
162
7-9
18
8
144
 
$\sum\text{f}_\text{i}=76$
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=412$
$\therefore$ Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{412}{76}=5.42$
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Question 192 Marks
The arithmetic mean of the following frequency distribution is 50.
Class
0-10
10-20
20-30
30-40
40-50
Frequency
16
p
30
32
14
Find the value of p.
Answer
Class
Frequency ($f_i$​​​​​​​)
Mid value ($x_i​​​​​​​$)
$(x_i \times f_i)$
0-10
16
5
80
10-20
pp
15
15pp
20-30
30
25
750
30-40
32
35
1120
40-50
14
45
630
 
$\sum\text{f}_\text{i}=92+\text{p}$
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=2580+15\text{p}$
Now, Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow25=\frac{2580+15\text{p}}{92+\text{p}}$
$\Rightarrow25(92+\text{p})=2580+15\text{p}$
$\Rightarrow2300+25\text{p}=2580+15\text{p}$
$\Rightarrow10\text{p}=280$
$\Rightarrow\text{p}28$
Note: Quastion modified
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Question 202 Marks
While calculating the mean of a given data by the assumed-mean method, the following values were obtained:
$\text{A}=25,\ \sum\text{f}_\text{i}\text{d}_\text{d}=110,\ \sum\text{f}_\text{i}=50.$
Find the mean.
Answer
$\text{A}=25,\ \sum\text{f}_\text{i}=50$ and $\sum\text{f}_\text{i}\text{d}_\text{i}=110$
Mean $=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=25+\frac{110}{50}$
$=25+2.2$
$=27.2$
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Question 212 Marks
Find the median of following data:
Class interval
0-10
10-20
20-30
30-40
40-50
Total
Frequency
8
16
36
34
6
100
 
Answer
Class interval
Frequency ($f_i$)
Cumulative frequency (cf)
0-10
8
8
10-20
16
24
20-30
36
60
30-40
34
94
40-50
6
100
Here, $\text{N}=100\Rightarrow\frac{\text{N}}{2}=50$
The cumulative frequency just greater than 50 is 60.
Hence, median class is 20-30.
$\therefore\text{l}=20,\ \text{h}=10,\ \text{f}=36,\ \text{cf}=$ cf of preceding class = 24
Now, Median $=\text{l}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$=20+\Big\{10\times\frac{50-24}{36}\Big\}$
$=20+\Big\{10\times\frac{26}{36}\Big\}$
$=20+7.2$
$=27.2$
Thus, the median of the data is $27.2.$
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